The super computer at Ram Mohan Roy Seminary takes an input of a number N and a X where X is a factor of the number N. In a particular case N is equal to 83p796161q and X is equal to 11 where 0

• It says that N is divisible by X. Therefore as X is 11, the no 83P796161Q to be divisible by 11 the difference of sum of digits in even place and the sum of digit in odd place should be equal o zero or divisible by 11. Therefore(8+p+9+1+1)-(3+7+6+6+q)=3+q-p should be divisible by 11. (cannot be zero as q>p.
  as q and p are both digits so 3+q-p can only be 11. Thus if q is 9 then p is 1, if q is 8 p is 0. as p>0. the only solution is 1=9, p=1.
  therefore the required no is 8317961619 which when divided by 9+1=10 gives a remainder of 9.
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